Operational Amplifier Circuits
Review:
Ideal Op-amp in an open loop configuration
Ip
+
Vp
+
Vi
_
Vn
Ro
+
Ri
Vo
AVi
_
In
An ideal op-amp is characterized with infinite open–loop gain
A→∞
The other relevant conditions for an ideal op-amp are:
1. Ip = In = 0
2. Ri = ∞
3. Ro = 0
Ideal op-amp in a negative feedback configuration
When an op-amp is arranged with a negative feedback the ideal rules are:
1. Ip = In = 0 : input current constraint
2. Vn = Vp : input voltage constraint
These rules are related to the requirement/assumption for large open-loop gain
A → ∞ , and they form the basis for op-amp circuit analysis.
The voltage Vn tracks the voltage Vp and the “control” of Vn is accomplished via the
feedback network.
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Page 1
Operational Amplifier Circuits as Computational Devices
So far we have explored the use of op amps to multiply a signal by a constant. For the
inverting amplifier the multiplication constant is the gain − RR12 and for the non inverting
amplifier the multiplication constant is the gain 1+ RR12 . Op amps may also perform other
mathematical operations ranging from addition and subtraction to integration,
differentiation and exponentiation.1 We will next explore these fundamental
“operational” circuits.
Summing Amplifier
A basic summing amplifier circuit with three input signals is shown on Figure 1.
Vin1
Vin2
Vin3
R1
I1
R2
I2
R3
I3
IF
RF
N1
Vout
Figure 1. Summing amplifier
Current conservation at node N1 gives
I1 + I 2 + I 3 = I F
(1.1)
By relating the currents I1, I2 and I3 to their corresponding voltage and resistance by
Ohm’s law and noting that the voltage at node N1 is zero (ideal op-amp rule) Equation
(1.1) becomes
Vin1 Vin 2 Vin 3
V
+
+
= − out
R1 R 2 R3
RF
(1.2)
1
The term operational amplifier was first used by John Ragazzini et. al in a paper published in 1947. The
relevant historical quotation from the paper is:
“As an amplifier so connected can perform the mathematical operations of arithmetic and calculus on the
voltages applied to its inputs, it is hereafter termed an ‘operational amplifier’.”
John Ragazzini, Robert Randall and Frederick Russell, “ Analysis of Problems in Dynamics by Electronics
Circuits,” Proceedings of IRE, Vol. 35, May 1947
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And so Vout is
RF
RF
⎛ RF
⎞
Vout = − ⎜
Vin1 +
Vin 2 +
Vin 3 ⎟
R2
R2
⎝ R1
⎠
(1.3)
The output voltage Vout is a sum of the input voltages with weighting factors given by the
values of the resistors. If the input resistors are equal R1=R2=R3=R, Equation (1.3)
becomes
Vout = −
RF
(Vin1 + Vin 2 + Vin3 )
R
(1.4)
The output voltage is thus the sum of the input voltages with a multiplication constant
RF
given by
. The value of the multiplication constant may be varied over a wide range
R
and for the special case when RF = R the output voltage is the sum of the inputs
Vout = − (Vin1 + Vin 2 + Vin 3 )
(1.5)
The input resistance seen by each source connected to the summing amplifier is the
corresponding series resistance connected to the source. Therefore, the sources do not
interact with each other.
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Difference Amplifier
This fundamental op amp circuit, shown on Figure 2, amplifies the difference between
the input signals. The subtracting feature is evident from the circuit configuration which
shows that one input signal is applied to the inverting terminal and the other to the noninverting terminal.
Vin1
Vin2
R1
I1
IF
R2
N1
R3
Vout
I2
R4
Figure 2. Difference Amplifier
Before we proceed with the analysis of the difference amplifier let’s think about the
overall behavior of the circuit. Our goal is to obtain the difference of the two input
signals (Vin 2 - Vin1 ) . Our system is linear and so we may apply superposition in order to
find the resulting output. We are almost there once we notice that the contribution of the
signal Vin2 to the output is
⎛ R4 ⎞ ⎛ R2 ⎞
Vout 2 = Vin 2 ⎜
⎟ ⎜1 +
⎟
R1 ⎠
⎝ R3 + R 4 ⎠ ⎝
(1.6)
and the contribution of signal Vin1 is
⎛ R2 ⎞
Vout1 = -Vin1 ⎜
⎟
⎝ R1 ⎠
(1.7)
R2
⎛ R4 ⎞ ⎛ R2 ⎞
Vout = Vout 2 - Vout1 = Vin 2 ⎜
⎟ ⎜1 +
⎟ - Vin1
R1 ⎠
R1
⎝ R3 + R 4 ⎠ ⎝
(1.8)
And the output voltage is
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Note that in order to have a subtracting circuit which gives Vout=0 for equal inputs, the
weight of each signal must be the same. Therefore
⎛ R4 ⎞ ⎛ R2 ⎞ R2
⎜
⎟ ⎜1 +
⎟=
R1 ⎠ R1
⎝ R3 + R 4 ⎠ ⎝
(1.9)
R4 R2
=
R3 R1
(1.10)
R2
(Vin 2 - Vin1 )
R1
(1.11)
which holds only if
The output voltage is now
Vout =
which is a difference amplifier with a differential gain of R2/R1 and with zero gain for
the common mode signal. It is often practical to select resistors such as R4=R2 and
R3=R1.
The fundamental problem of this circuit is that the input resistance seen by the two
sources is not balanced. The input resistance between the input terminals A and B, the
V
differential input resistance, Rid (see Figure 3) is Rid ≡ in
I
A
+
V in
B
I
R1 V -
R3 V +
R2
Vout
R4
Figure 3. Differential amplifier
Since V+ = V- , Vin = R1 I + R3 I and thus Rid = 2 R1 . The desire to have large input
resistance for the differential amplifier is the main drawback for this circuit. This problem
is addressed by the instrumentation amplifier discussed next.
Instrumentation Amplifier
Figure 4 shows our modified differential amplifier called the instrumentation amplifier
(IA). Op amps U1 and U2 act as voltage followers for the signals Vin1 and Vin2 which see
the infinite input resistance of op amps U1 and U2. Assuming ideal op amps, the voltage
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Page 5
at the inverting terminals of op amps U1 and U2 are equal to their corresponding input
voltages. The resulting current flowing through resistor R1 is
V −V
(1.12)
I1 = in1 in 2
R1
Since no current flows into the terminals of the op amp, the current flowing through
resistor R2 is also given by Equation (1.12).
Vin1
V01
U1
V1
R4
R2
R3
R1
U3
R3
Vout
R2
R4
V2
V02
U2
Vin2
Figure 4. Instrumentation Amplifier circuit
Since our system is linear the voltage at the output of op-amp U1 and op-amp U2 is
given by superposition as
R2
⎛ R2 ⎞
V01 = ⎜ 1 +
Vin2
⎟ Vin1 −
R1 ⎠
R1
⎝
(1.13)
R2
⎛ R2 ⎞
V02 = ⎜ 1 +
Vin1
⎟ Vin2 −
R1 ⎠
R1
⎝
(1.14)
Next we see that op amp U3 is arranged in the difference amplifier configuration
examined in the previous section (see Equation (1.11)). The output of the difference
amplifier is
Vout =
The differential gain,
R4 ⎛ 2R2 ⎞
⎜1 +
⎟ ( Vin2 − Vin1 )
R3 ⎝
R1 ⎠
(1.15)
R4 ⎛
2R2 ⎞
⎜1 +
⎟ , may be varied by changing only one resistor: R1.
R3 ⎝
R1 ⎠
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Current to voltage converters
A variety of transducers produce electrical current in response to an environmental
condition. Photodiodes and photomultipliers are such transducers which respond to
electromagnetic radiation at various frequencies ranging from the infrared to visible to γrays.
A current to voltage converter is an op amp circuit which accepts an input current and
gives an output voltage that is proportional to the input current. The basic current to
voltage converter is shown on Figure 5. This circuit arrangement is also called the
transresistance amplifier.
R
N1
Vout
I in
Figure 5. Current to voltage converter
Iin represents the current generated by a certain transducer. If we assume that the op amp
is ideal, KCL at node N1 gives
⎛V −0 ⎞
I1 + ⎜ out
⎟ = 0 ⇒ Vout = − RI1
⎝ R ⎠
(1.16)
The “gain” of this amplifier is given by R. This gain is also called the sensitivity of the
converter. Note that if high sensitivity is required for example 1V/µV then the resistance
R should be 1 MΩ. For higher sensitivities unrealistically large resistances are required.
A current to voltage converter with high sensitivity may be constructed by employing the
T feedback network topology shown on Figure 6.
In this case the relationship between Vout and I1 is
⎛ R2 R2 ⎞
Vout = − ⎜ 1 +
+
⎟ I1
R1 R ⎠
⎝
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(1.17)
Page 7
R
N1
R1
R2
Vout
I in
Figure 6. Current to voltage converter with T network
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Voltage to Current converter
A voltage to current (V-I) converter accepts as an input a voltage Vin and gives an output
current of a certain value.
In general the relationship between the input voltage and the output current is
I out = SVin
(1.18)
Where S is the sensitivity or gain of the V-I converter.
Figure 7 shows a voltage to current converter using an op-amp and a transistor. The opamp forces its positive and negative inputs to be equal; hence, the voltage at the negative
input of the op-amp is equal to Vin. The current through the load resistor, RL, the
transistor and R is consequently equal to Vin/R. We put a transistor at the output of the
op-amp since the transistor is a high current gain stage (often a typical op-amp has a
fairly small output current limit).
Vcc
RL
V in
R
Figure 7. Voltage to current converter
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Amplifiers with reactive elements
We have seen that op amps can be used with negative feedback to make simple linear
signal processors. Examples include amplifiers, buffers, adders, subtractors, and for each
of these the DC behavior described the apparent behavior over all frequencies. This of
course is a simplification to treat the op amp ideally, as through it does not contain any
reactive elements. Providing we keep the operating conditions out of the slew rate limit
then this is a reasonable model. Here we wish to extend this picture of op amp operation
to include circuits that are designed to be frequency dependent. This will enable the
construction of active filters, integrators, differentiators and oscillators.
The feedback network of an op-amp circuit may contain, besides the resistors considered
so far, other passive elements. Capacitors and inductors as well as solid state devices
such as diodes, BJTs and MOSFETs may be part of the feedback network.
In the general case the resistors that make up the feedback path may be replaced by
generalized elements with impedance Z1 and Z2 as shown on Figure 8 for an inverting
amplifier.
Z2
Z1
Vout
Vin
Figure 8. Inverting amplifier with general impedance blocks in the feedback path.
For an ideal op-amp, the transfer function relating Vout to Vin is given by
Z (ω )
Vout
=− 2
Vin
Z1 (ω )
(1.19)
Now, the gain of the amplifier is a function of signal frequency (ω) and so the analysis is
to be performed in the frequency domain. This frequency dependent feedback results in
some very powerful and useful building blocks.
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The Integrator: Active Low Pass Filter
The fundamental integrator circuit (Figure 9) is constructed by placing a capacitor C, in
the feedback loop of an inverting amplifier.
C
IC
R
V in
V out
IR
R
Figure 9. The integrator circuit
Assuming an ideal op-amp, current conservation at the indicated node gives
I R = IC
(1.20)
Vin
dV
= −C out
R
dt
Rearranging Equation (1.20) and integrating from 0 to t, we obtain
∫ dVout = − ∫
Vin (τ )
dτ
RC
t
⇒ Vout (t ) = −
1
Vin (τ ) dτ + Vout (0)
RC ∫0
(1.21)
The output voltage is thus the integral of the input. The voltage Vout (0) is the constant of
integration and corresponds to the capacitor voltage at time t = 0.
The frequency domain analysis is obtained by expressing the impedance of the feedback
components in the complex plane. The transfer function may thus be written as
1
Vout
Z
j
jωC
=− C =
=
Vin
ZR
R
ω RC
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(1.22)
Page 11
The above expression indicates that there is a 90o phase difference between the input and
the output signals. This 90o phase shift occurs at all frequencies. The gain of the amplifier
V
1
is also a function of frequency. For dc signals with
given by the modulus out =
Vin
ω RC
ω=0 the gain is infinite and it falls at a rate of 20dB per decade of frequency change. The
infinite gain for dc signals represents a practical problem for the circuit configuration of
Figure 27. Since the equivalent circuit of a capacitor for ω=0 is an open circuit, the
feedback path is open. This lack of feedback results in a drift (cumulative summing) of
the output voltage due to the presence of small dc offset voltages at the input. This
problem may be overcome by connecting a resistor, RF, in parallel with the feedback
capacitor C as shown on Figure 10.
C
IC
RF
R
V in
V out
IR
R
Figure 10. Active Low Pass filter
The feedback path consists of the capacitor C in parallel with the resistor RF. The
equivalent impedance of the feedback path is
RF
R Ζ
RF
jωC
ZF = F C =
=
1
1 + jω RF C
RF + ΖC R +
F
jωC
The transfer function
(1.23)
Z (ω )
Vout
becomes
=− F
Vin
Z1 (ω )
Z (ω )
Vout
R
R
1
1
=− F
=− F
=− F
Vin
Z1 (ω )
R1 1 + jω RF C
R1 1 + jω
(1.24)
ωΗ
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Where
ωH ≡
1
RF C
(1.25)
Vout
versus frequency. At frequencies much less
Vin
R
than ωH (ω << ωH) the voltage gain becomes equal to F , while at frequencies higher
R
than ωH (ω >> ωH) the gain decreases at a rate of 20dB per decade.
Figure 11 shows the logarithmic plot of
Figure 11. Bode plot of active low pass filter with a gain of 5.
So we have seen that the integration is achieved by charging the feedback capacitor.
For an integrator to be useful it must be allowed to be reset to zero. Since the output is
stored in the charge of the feedback capacitor, all we need to do is to short out the
capacitor in order to reset the integrator.
Integrators are very sensitive to DC drift, small offsets lead to a steady accumulation of
charge in the capacitor until the op amp output saturates. We can avoid this by providing
another feedback path for DC. The circuit incorporates a shorting path across the
capacitor as shown on Figure 12.
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Page 13
C
IC
RF
R
V in
V out
IR
R
Figure 12. Integrator with reset button
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The Differentiator: Active High Pass Filter
A differentiator circuit may be obtained by replacing the capacitor with an inductor for
Figure 9. In practice this is rarely done since inductors are expensive, bulky and
inefficient devices. Figure 13 shows a fundamental differentiator circuit constructed with
a capacitor and a resistor.
R
C
V out
V in
R
Figure 13. The differentiator circuit
For an ideal op-amp, the current flowing through the capacitor, C
current flowing through the resistor,
dVin
, is equal to the
dt
Vout
, and thus
R
Vout = − RC
dVin
dt
(1.26)
The output is thus proportional to the derivative of the input.
As the integrator is sensitive to DC drifts, the differentiator is sensitive to high frequency
noise. The differentiator thus is a great way to search for transients, but will add noise.
The integrator will decrease noise. Both of these arguments assume the common situation
of the noise being at higher frequency than the signal.
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Active Band Reject Filter
The integrator and differentiator demonstrate that op amp circuits can be designed to be
frequency dependent. This permits the design of active filters, a filter that has gain. We
saw before that we could design passive filters based on LC circuits, active filters are no
more complicated. Simple selective filters can be made through a frequency dependent
impedance in the feedback loop.
Consider the band reject circuit shown on Figure 14.
C
L
RF
IF
R1
V in
V out
IR
R
Figure 14. Active band reject filter
We can understand how this circuit works without any detailed calculations. All we need
to do is look at the feedback loops. There are two paths in the feedback loop: a frequency
independent path with resistance RF, and a frequency dependent path with impedance
given by
1 ⎞
⎛
Zω = j ⎜ ω L −
ωC ⎟⎠
⎝
(1.27)
Let’s look at the behavior of the circuit as a function of frequency.
For DC signals ( ω = 0 ) the capacitor acts as an open circuit and the equivalent circuit is
shown on Figure 15.
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RF
IF
R1
V in
V out
IR
R
Figure 15
Similarly at high frequencies ( ω → ∞ ) the inductor acts as an open circuit and the
equivalent circuit is the same as the one shown on Figure 15.
Therefore the voltage transfer characteristics at DC and at high frequency are the same
with a gain given by
G=
Vout
R
=− F
Vin
R1
(1.28)
The other frequency of interest if the resonance frequency, which occurs when Zω as
given by Equation (1.27) is equal to zero. The resonance frequency is
1
LC
and the circuit reduces to the one shown on Figure 16.
ωo =
(1.29)
IF
R1
V in
V out
IR
R
Figure 16
which gives Vout=0 at ω = ωo .
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Page 17
So this is a filter that passes and amplifies every frequency except the resonance
frequency.
For the full analysis of this active filter we may write down the complete expression for
the impedance of the feedback loop which is
1 ⎞
⎛
j ⎜ωL −
R
j (ω 2 LC − 1) RF
ωC ⎟⎠ F
⎝
Z F = Zω // RF =
=
1 ⎞ ω RF C + j (ω 2 LC − 1)
⎛
RF + j ⎜ ω L −
ωC ⎟⎠
⎝
(1.30)
And thus the general transfer function of the filter is
j (ω 2 LC − 1) RF
Vout
ZF
=−
=
Vin
R1 R1 ⎡ω RF C + j (ω 2 LC − 1) ⎤
⎣
⎦
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(1.31)
Page 18
Diodes and transistors in op-amp circuits.
Diodes and transistors may also be used in op-amp circuits. The nonlinear behavior of
these devices result in very interesting and useful non-linear op-amp circuits.
Logarithmic Amplifier
If we are interested in processing a signal that has a very wide dynamic range we take
advantage of the exponential i-v characteristics of the diode and design an amplifier
whose output is proportional to the logarithm of the input.
In practice we may have the voltage signal that corresponds to a certain chemical activity
such as the activity, or concentration, of hydrogen ions in a solution which represents the
pH of the solution. In this case the voltage is exponentially related to the concentration
(pH).
Vi = Vo k ln( pH )
(1.32)
If we use this signal as the input to an inverting amplifier we may linearize the signal by
using a diode in the feedback path of the amplifier.
Recall the i-v relationship for a diode
I = Io ⎡⎣ eqV / kT − 1⎤⎦
IoeqV / kT
(1.33)
Consider the circuit shown on Figure 17.
I
R1
IR
Vi
Vn
Vo
Vp
R
Figure 17. Logarithmic Amplifier
KCL at the indicated node gives
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Vi − Vn
= Ioe q (Vn −Vo ) / kT
R1
(1.34)
Vi
= Ioe q ( −Vo ) / kT
R1
(1.35)
And since Vn = Vp = 0 we obtain
And solving for Vo gives the desired relationship.
Vo = −
kT
Vi
ln
q
Io R1
Vo = −
kT
kT
ln(Vi ) +
ln( IoR1)
q
q
N
a
(1.36)
b
Vo = − a ln(Vi ) + b
Similarly and antilogarithmic amplifier may be constructed by placing the diode in
series with the signal source as shown on Figure 18.
R2
IF
Vi
Vn
I
Vo
Vp
R
Figure 18. Antilogarithmic amplifier
Here you may show that
Vo = − IoR 2e qVi / kT
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(1.37)
Page 20
Superdiode. Precision half wave Rectifier
The diode rectifier circuit and its associated voltage transfer characteristic curve are
shown on Figure 19(a) and (b).
Vout
D
Vin
+ Vd R11
Vout
Vd
Vin
(b)
(a)
Figure 19. Diode rectifier circuit (a) and voltage transfer curve (b)
The offset voltage Vd is about 0.7 Volts and this offset value is unacceptable in many
practical applications. The operational amplifier and the diode in the circuit of Figure 20
form an ideal diode, a superdiode, and thus they eliminate the offset voltage Vd from the
voltage transfer curve forming an ideal half wave rectifier.
Superdiode
Vout
V-
I1
V1
V+
Vin
D
Id + Vd -
I2
R
Vout
Vin
Figure 20. Precision half wave rectifier circuit and its voltage transfer curve.
Let’s analyze the circuit by considering the two cases of interest: Vin>0 and Vin<0.
For Vin<0 the current I2 and id will be less than zero (point in a opposite direction to the
one indicated). However, negative current can not go through the diode and thus the
diode is reverse biased and the feedback loop is broken. Therefore the current I2 is zero
and so the output voltage is also zero, Vout=0. Since the feedback loop is open the
voltage V1 at the output of the op-amp will saturate at the negative supply voltage.
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For Vin>0, Vout=Vin and the current I2=Id and the diode is forward biased. The
feedback loop is closed through the diode. Note that there is still a voltage drop Vd across
the diode and so the op-amp output voltage V1 is adjusted so that V1=Vd+Vin.
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Problems
P1. Resistors R1 and R2 of the circuit on Figure P1 represent two strain gages placed
across each other on a beam in order to measure the tensile and compressive stains. R1
and R2 vary symmetrically by a factor δ as R1=R(1-δ) and R2=R(1+δ).
Vb
R1
+
_
R
Vs
R2
R
Figure P1.
Design an amplifier so that the output varies from -10V to +10V as the parameter δ varies
from -0.01 to +0.01. The bias voltage Vb=+10V and R=10kΩ.
P2. The resistors of the amplifier circuit of Figure P2 have a tolerance of ±δ%.
1. Assume that Vin is known precisely and calculate the deviation in the output
voltage Vout.
2. For R1=15kΩ±5% and R2=200kΩ±5% and Vin=100mV±1% calculate the output
voltage.
R2
R1
V out
V in
Figure P2.
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P3. Calculate the output voltage Vout for the following circuits.
V in
V out
R
3R
2R
R2
V in
R2
V out
2R
3R
R
P4 For the circuit on Figure P4 determine the value of resistor Rx so that the output
voltage is zero.
Rx
R2
V out
V in
R3
R4
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P5 The circuit on Figure P5 is a current source.
1. Show that the amount of current delivered to the load is controlled by resistor R3.
2. Calculate the resistance seen by the Load.
R2
R1
V out
V in
R4
R3
Load
Figure P5
P6. For the circuit on Figure P6 calculate the currents i1, i2, i3, i4 and the voltages v1 and
Vout. Refer your answers to the indicated current directions.
R2
R1
R4
i2
R3
i1
v1
V in
i3
i4
V out
Figure P6
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P7
For the voltage to current converter, briefly (in 1-2 sentences) describe what happens
when each of the following faults happen (alone, independently of other faults).
a. the Zener diode is shorted
b. R5 has a bad soldering joint and is opened
c. The load becomes shorted
d. The connection between the op-amp output and the base of the transistors
becomes opened
P8: Op-amp nonidealities:
a. What is the effect of a 1uA input bias current on the output voltage of the opamp?
b. What is the effect of a 5mV input offset voltage on the output of the opamp?
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